The figure shows a bicycle wheel,which is rotating with large angular velocity omega about the axle AB,hung from a string connected to the end A and holding from end B.If it is released from end B,

(1) the end B will fall down and the axle AB becomes vertical.
(2) the direction of AB remains unchanged.
(3) tha axle will rotate about the vertical axis through A,while AB remains approximately horizontal
(4) the end B will fall down and the wheel will start to oscillate like a pendulum.
(5) the end B will move upward first and then fall down and will start to oscillate like a pendulum.
(ans:3)


Well,my first choice was number (4),I just thought that once we let go of the rod ,the weight of the wheel+ rod would act vertically down and would overcome any othe force present(?) so the end B will fall down,and probably oscillate like a pendulum.(just a bad guess)
But,I guess this has alot to do with momentum etc,but I can't seem to figure out why it's (3) and why it seems to rotate about A.
Hope someone can help

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Well, when the rod is released, the free end tends to fall due to its weight and then it is a case of simple pendulum motion. I don't the spinning of the wheel will affect the Harmonic Motion and so the wheel continues spinning about the axis.

Actually I think it'll move like a gyroscope but that's not relevant here.
The motion would be simple-pendulum like, I think.

(I think) I get what you're saying,but the answer is (3),
the axle will rotate about the vertical axis through A,while AB remains approximately horizontal.
but how?

I feel Akshay is right and the Gyroscopic effect comes into the picture.
It is this effect which makes it easier to balance on a moving bicycle compared to a stationary one.

Thank you sir,but ,
this is the first time I've heard of a gyroscope(don't think it's in our syllabus,but may have learnt the same principles,otherwise it wouldn't be included in our pastpaper,)
I found a very helpful link on the gyrascopic effect,I understood most of it (I think?)upto the point they say a torque begins to act on the gyrascope,(which would mean a couple would act on this axle too,ryt?).
I don't get this,how does a torque affect the motion and why(and how) does it appear?
What is Gyroscopic Effect

I read another link which explained the gyrascope effect in alot of detail,but it only helped me get more confused,
gyroscope

Hope someone can help explain this effect for me,and this too,if possible.
Thank you.

Thanks for the links. In the first one they have given the example of a spinning top which doesnt fall, while a stationary one does. Aplly the same idea horizontally to a spinning bicycle wheel and you will understand.

Yes,you're right,the answer's right there in the 3rd para.
There's a torque acting through the spin axis and the axis of rotation which causes the rod to rotate around end A and, AB would approximately(I guess cause of the effect of the weight of rod) remain horizontal.
ok,i understand this but, one things still unclear to me,
what causes this torque?
is it due to the tension in the string and the weight of the system?
I would really appreciate any help.
Thank you

Pls may you explain the logic of the answer..

I am also new to this kind of motion. i also thought the pendulum option was correct.
but going through the links, i understood from where the torque is originating.
initially, when rod is held , weight of wheel plus rod is balanced by upward tension and normal contact force. once hand is removed, if i consider the point where string makes contact with the rod, the torque about this point is(r cross w) where r= distance of com from the point considered.w=weight,.The vector cross product has direction along the tangent. applying tow= dL/dt, the change in angular mom must be along tangent, the only way to perform this is that the axle should rotate, otherwise how will L change, that too along the tangent?
one thing puzzled me though, this peculiar motion is also creating angular momentum about z axis, strangely it is not considered at all !